C hem F actsheet September 2002
Number 40
Reaction Mechanisms To succeed in this topic you need to: Have a good knowledge of the AS and A2 level organic chemistry reactions covered so far; Be confident in using organic terminology, terminology, nomenclature and structural • formulae.
Exam Hint: - Reaction
mechanisms are commonly examined. As well as learning the standard exampl es given here, you need to practise applying the mechanisms to other similar reactions.
•
After working through this Factsheet you will: Be able to represent movement of electron using the “curly arrow” convention; Have been shown the key reaction mechanisms. •
2. Homolytic Free Radical Addition.
•
Example: Polymerisation of ethene. The initiator is a free radical, R . (which could be a peroxide radical from the oxygen catalyst).
All chemical reactions involve the movement of electrons, and reaction mechanisms allow us to display these movements and understand more about how a reaction occurs.
Stage 1 - Initiation. H R . H
In this Factsheet you will meet organic reactions which you have come across before. You should already be able to recall the overall reaction equations and conditions required. The reaction mechanisms, which also require learning, often split these reactions up into several steps, each step involving the movement of elec trons. This should give an insight into how a reaction occurs.
C
H H
C
R
Stage 3 - Termination. H H H
represents the movement of a single electron.
n
C .
C
.C
H C
n
n
These reactions involve free radicals, radicals, very reactive species which have a single, unpaired electron.
Cl .
.Cl
Cl
Cl
Example: Alkanes with chlorine.
Cl .
.CH 3
Cl
CH 3
Stage 1 - Initiation
CH4 + Cl2
H Cl
Cl
Cl
C . + HCl
H3C
uv light
CH3 + CH3
H
H
C
C
C
C
H radical
n
CH3Cl + HCl
C 2H 6
Most free radical radical reactions are extremely unstable and explosive. This reaction between methane and chlorine is controlled using subdued UV light.
H
H H 3 C .
H
H
C .
But be aware that “by-products” are also formed in such reactions, e.g. the formation of ethane (when chloromethane is the target molecule).
Stage 2 - Propagation
H
H
H
The overall reaction equation is as follows:
Cl . + Cl .
The Cl-Cl bond is broken homolytically to form two chlorine free radicals. Energy is required to break the bond, hence the reaction only begins in the presence of UV light.
C
H
Free radicals are used up and not regenerated, causing the reaction to eventually stop.
The reaction between methane and chlorine occurs in 3 stages.
H
H
Stage 3 - Termination
1. Homolytic Free Radical Substitution
H
C .
H H H H H H H H The end product, polyethene, consists of long chain molecules (of varying length) which are branched.
When using curly arrows it is vital that you position each end of the arrows carefully, so that you clearly show the examiner that you know where the electrons are moving from and to. to.
uv light
C
Stage 2 - Propagation. H H H H H H H R C C C C C C C . R H H H H H H H Termination can occur when two radical chains meet, or when a chain attacks an established chain.
represents the movement of an electron pair
Cl
H
The new radical formed can then react with further ethene molecules to lengthen the polymer chain.
You need to be able to represent the movement of electrons using the “curly arrow” convention:
Cl
H
Cl + Cl .
Any such reaction caused by a free radical, where “replacement” free radicals are produced by the reaction, hence the reacti on process continues unaided.
1
Reaction mechanisms
Chem Factsheet
Note the formation of a carbocation (C + ) intermediate. The most stable position for the carbocation will be the carbon with the least hydrogen atoms around it.
3. Heterolytic Electrophylic Addition Example: Alkenes with halogens or hydrogen halides. Consider the addition of bromine to propene. H H H3C C C H δ+ Br
hence
CH 3
Br
Br
Br
4. Heterolytic Electrophilic Electrophilic Substitution Substitution
conc. H 2SO4
NO 2 +
0
50 C
H C
C δ+
H
δ−
Br
H 2O
+ H3 O
Br
Br
H
H
+
−
FeBr4 Br + H
+
−
+ FeBr4
HBr
+
+ FeBr3
AlCl 3
+ HCl
The mechanism is as follows: H H C Cl AlCl3
+
CH 2CH 3 H
CH 3
−
+ AlCl4
CH 2CH 3 + HCl
+ AlCl3
ethylbenzene A similar reaction mechanism occurs when benzene reacts with acid chlorides and AlCl3 to give ketones. O O
+ NO2
+ H
H
HBr
Br H
+
+ C2H5Cl
H
NO 2
Br
H
(c) Alkylation of benzene with with chloroalkanes or acyl acyl chlorides Benzene reacts with chloroalkanes in the presence of anhydrous aluminium chloride to give an alkylbenzene. C 2H5
+
or
C
Bromobenzene is formed and the H + is the removed by FeBr4 − to regenerate the FeBr3.
The intermediate cation will then break down to either reform benzene, or form nitrobenzene. The delocalised system reforms, hence energy is released.
+
H
C
Bromobenzene
Note that a considerable amount of energy is required to disrupt the very stable π-system, therefore this step has a very high activation energy.
NO 2
H → CH 3
H
+
NO 2
+
→
+
+
NO 2+ then attacks a C atom, forming a bond and distrupting the delocalised π-system. NO 2 +
Br
FeBr 3
Br
First a “loose” association is formed.
NO 2
C
+
+
Then:
The NO2+ ion is a strong electrophile, so attacks the delocalised πelectron system in the benzene ring.
→
C
H
2Fe + 3Br2 → 2FeBr3
Note that the nitric acid is acting as a base (proton acceptor) in the presence of a stronger acid.
+
→ CH 3
H
The electrophile Br+ must be generated to attack the benzene ring.
Evidence suggests that this is the NO2 species - an electrophile formed by the removal of OH − from HNO3. HNO3 + 2H2SO4 → NO2 + 2HSO
H
H
Fe cat.
+
+ NO2
Br
−
Br2 +
Therefore, sulphuric acid must react with nitric acid forming a species which will attack the benzene ring.
4
H
+
Br
Nitrobenzene The reaction of benzene with nitric acid alone is slow and pure sulphuric acid at 500C has little or no effect on benzene.
+
C
(b) Bromination of benzene Bromine reacts with benzene in the presence of an iron wire catalyst in dry conditions at room temperature.
Examples: Reactions of benzene (see Chem Factsheet 39). (a) Nitration of benzene Benzene reacts with a mixture of conc. nitric acid and conc. sulphuric acid at 50oC to form nitrobenzene. + HNO3
C
Markownikov’s Rule When a hydrogen halide is added to a C=C double bond, the hydrogen atom is added to the carbon atom that already carri es more hydrogens.
+
Br
H was formed, not CH 3
H
:Br
Meanwhile, the δ− end of the halogen molecule will accept both of the bond pair of electrons, forming a negative anion . H H H H H H CH 3 C C H → CH 3 C C H → CH 3 C C H −
C
+
H
The δ+ end of the halogen molecule is electrophilic, and can attack the electron rich π-bond.
:Br
C
H
There is a similar reaction mechanism for the addition of hydrogen halides to alkenes.
The electrons in the double bond repel the electrons in the approaching Br2 molecule, causing a dipole.
Br
H
Br
Br
δ−
CH 3
H
+ CH 3
+
AlCl 3
C Cl
Nitrobenzene
2
C
+ HCl Cl
Reaction mechanisms
Chem Factsheet
5. Heterolytic Nucleophilic Substitution Example: Halogenoalkanes with hydroxide ions and cyanide ions.
(2) SN2 mechanism The mechanism involves the substitution of a nucleophile which is second order (two molecules in rate determining step).
This mechanism is associated with polar covalent bonds (e.g. carbon-halogen). δ+
C
The nucleophile attacks the Cδ+ and starts to form a covalent bond with it. At the same time the C δ+-Halδ− bond starts to break heterolytically.
δ−
Br
The positive charge on the carbon allows attack by nucleophiles such as :OH− or :CN− which have lone pairs to donate.
This is one continuous process with the carbon-nucleophile bond getting stronger and the carbon-halogen getting weaker.
There are two mechanisms for nucleophilic substitution, (1) SN1 and (2) SN2.
The “transition state” (corresponding to the activation energy peak) is when both are of equal strength. CH 3 CH 3 CH
(1) SN1 mechanism This mechanism involves the substitution of a nucleophile which is first order (one molecule in rate determining step).
-
3
HO :
Stage 1: (The rate determining step) CH 3 CH 3 CH 3 C
Br
CH 3 Stage 2: CH 3 CH 3 C
+ +
+ + Br -
SN1 or SN2? To find out which mechanism is followed follow ed for such nucleophilic substitution reactions, the reaction rate must be followed by experiment and the rate order with respect to the nucleophile discovered.
OH
CH 3
In general the mechanism depends on the stability of the carbocation.
The first stage is rate determining as the second stage occurs much faster. Therefore the rate equation is:
Alkyl groups are more “electron donating” than hydrogen atoms, so the carbocation of a tertiary group is more stable than that of a primary group. H R1 R1
Rate = k[(CH3)3CBr] The reaction is zero order with respect to the nucleophile (OH − ) as it does not take part in the rate determining step.
R
This mechanism involves the formation of an intermediate - the carbocation.
C +
C +
R2
H Primary
R2
favour SN1
Answers
6. Heterolytic Nucleophilic Addition Example: Carbonyl compounds and hydrogen cyanide.
1. Homolytic Homolytic fission fission is is when a bond bond breaks breaks with one electron electron going going to each of the bonded atoms forming two free free radical species. Heterolytic fission is when a bond breaks with both of the bond pair electrons going to one of the bonded ato ms, forming a cation and anion.
A nucleophile will attack the δ+ carbon in the C=O group. δ+
δ−
C
O
R
R
+ C
R
-
O
:CN
-
- : O
R
H+
R
R
OH
2. (a) HNO3 + 2H 2SO4 → NO2+ + 2HSO4− + H 3O+
C
C CN
C +
R3 Tertiary
H Secondary
favour SN2
R
H
H
Rate = k[CH 3CH2Br][OH − ]
CH 3 C
CH 3
-
The formation of the transition state represents the rate determining step. This step is first order with respect to the halogenoalkane and the nucleophile.
CH 3
:OH
: Br
C
HO
Br
H H transition state
CH 3
-
C
HO
Br
H H
This mechanism occurs in two stages:
CH 3 C
C
R
+
CN
+
+ NO2 → (b)
Questions. 1. What is this difference between homolytic and heterolytic fission fission of bonds?
C 2H 5
H
H C
C
H
→
NO 2
→ C 2H 5
H
+
H
H
C
C
+
−
NO 2 H →
NO 2
H → C2H 5
H
+ H
H
H
C
C
Br
H
+
H
:Br
2. Show the the reaction reaction mechanis mechanisms ms for for the following following reaction reactions: s: (a) C6H6 + HNO3 → C6H5NO2 + H2O
Br (c)
(b) C2H5CHCH2 + HBr → C2H5CHBrCH3
C 2H 5 CH 3
(c) C2H5COCH3 + HCN → C2H5C(OH)CH3CN
δ+
C
C2H5 + - C2H 5 O : O C O → → C CH 3 CH 3 CN :CN δ−
H+
OH
C2H5 C
Acknowledgements: CH 3 CN This Factsheet was researched and written by Kieron Heath Curriculum Press, Unit 305B, The Big P eg, 120 Vyse Street, Street, Birmingham, B18 6NF ChemistryFactsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. ISSN 1351-5136 No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN
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